📚 CBSE Grade-10 Study Guide
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Mathematics — Standard (041) — CBSE Class 10 board question

Q1. [2] § 8.3 Trigonometric Ratios of Some Specific Angles § 8.5 Summary
Evaluate: $\dfrac{3\cos^2 30^\circ - 6\operatorname{cosec}^2 30^\circ}{\tan^2 60^\circ}$
Previously asked in CBSE board exam
2026 30/4/1 Q23 (OR-2)
Generated by claude-sonnet-4-6 · 2026-06-15 10:21 · grounding rag
Model Answer

Known values: $\cos 30° = \dfrac{\sqrt{3}}{2}$, $\operatorname{cosec} 30° = 2$, $\tan 60° = \sqrt{3}$

$$= \frac{3\cos^2 30° - 6\operatorname{cosec}^2 30°}{\tan^2 60°}$$

$$= \frac{3\left(\dfrac{\sqrt{3}}{2}\right)^2 - 6(2)^2}{(\sqrt{3})^2}$$

$$= \frac{3 \times \dfrac{3}{4} - 6 \times 4}{3}$$

$$= \frac{\dfrac{9}{4} - 24}{3} = \frac{\dfrac{9 - 96}{4}}{3} = \frac{-87}{4 \times 3} = \frac{-87}{12} = \mathbf{-\dfrac{29}{4}}$$

Source: Chapter 8, Section 8.3 (Trigonometric Ratios of Specific Angles)

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Explanation
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